Basic integration identity- please jog my memory

AI Thread Summary
The discussion centers on verifying that the functions f_{n}(x) = A cos(2πn x/L) can serve as basis functions for the interval 0<x<L. The user has successfully normalized the function but is struggling with checking orthogonality. To demonstrate orthogonality, the integral condition ∫ f_{n}(x) f_{m}(x)dx = δ_{n,m} must be satisfied. A suggested approach involves using the trigonometric identity cosAcosB = 1/2 [cos(A+B) + cos(A-B)] to simplify the integration of the product of two cosines. This identity will help in resolving the integration challenge and confirming the orthogonality condition.
Oscur
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Hello all, I'm doing a question for the maths module in my physics degree (I'm a second year undergrad) and there's a question I'm doing on basis functions.

Homework Statement



Verify that functions of the type f_{n}(x) = A cos \frac{2\pi n x}{L} where n = 0,1,2... can be used as a basis function for 0<x<L.

There are then three parts:

i. Normalise (I've done this part, I found a normalising factor of A=\sqrt{\frac{2}{L}}

ii. Check orthogonality (this is the part causing the trouble)

iii. Would it be possible to use f_{n}(x) as the basis for all functions in this interval? (this clearly follows from the other two parts, so once I've done ii, I should be able to answer this)

Homework Equations



To check orthogonality, I need to demonstrate that the condition:

\int f_{n}(x) f_{m}(x)dx=\delta_{n,m}, where the limits of the integral are the same as the interval of interest, is met.


The Attempt at a Solution



This means I need to integrate the product of two cosines with different arguments, and if I try and do this with integration by parts, I end up in an infinite loop with products of cyclic functions. Is there an identity for this?

Thanks in advance!

Matt
 
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Hello Matt! :smile:
Oscur said:
This means I need to integrate the product of two cosines with different arguments … Is there an identity for this?

It's one of the standard trigonometric identities …

cosAcosB = 1/2 [cos(A+B) + cos(A-B)] :wink:
 
Much appreciated, thank you!
 
Try using a trig identity to write cos(nx)cos(mx) in terms of cos((n-m)x) and cos((n+m)x).
 
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